16
Simulation
LEARNING OBJECTIVES :
After studying this chapter, you should be able to :
l
explain the term simulation and reasons for using simulation;
l
identify the steps in the simulation process;
l
review the reasons why simulation may be preferable to other decision models;
l
simulate a situation based on real data;
l
make useful contribution in (investment decision or capital budgeting) through
simulation;
l
establish how simulation permits study of a system under controlled conditions.
16.1 INTRODUCTION
For various managerial problems discussed so far, we have been able to find a mathematical
solution. However, in each of these cases the problem (e.g. linear programming or CPM/
PERT) was simplified by certain assumptions so that the appropriate mathematical
techniques could be employed. There are certain managerial situations which are so
complex that mathematical solution is impossible given the current state of the art in
mathematics. In many other cases, the solutions which result from simplifying assumptions
are not suitable for the decision makers. In these cases, simulation offers a reasonable
alternative.
16.2 WHAT IS SIMULATION?
Simulation is a quantitative procedure which describes a process by developing a model of
that process and then conducting a series of organised trial and error experiments to predict
the behavior of the process over time. Observing the experiments is much like observing
the process in operation. To find how the real process would react to certain changes, we
can introduce these changes in our model and simulate the reaction of the real process to
them. For example, in designing an airplane, the designer can build a scale model and
observe its behavior in a wind tunnel. In simulation, we build mathematical models which
we cannot solve and run them on trial data to simulate the behavior of the system.
16.2
Advanced Management Accounting
16.2.1 Steps in the simulation process : All simulations vary in complexity from situation
to situation. However, in general, one would have to go through the following steps:1.
Define the problem or system you intend to simulate.
2.
Formulate the model you intend to use.
3.
Test the model, compare its behavior with the behavior of the actual problem
environment.
4.
Identify and collect the data needed to test the model.
5.
Run the simulation.
6.
Analyze the results of the simulation and, if desired, change the solution you are
evaluating.
7.
Rerun the simulation to test the new solution.
8.
Validate the simulation, that is, increase the chances that any inferences you draw
about the real situation from running the simulation will be valid.
It may be noted that the fundamental principle of simulation is to make use of some device
that can represent the phenomenon of a real-life system to enable us to understand the
properties, behavior and functional characteristics of the system. The device used can be
any convenient means-a mathematical formula or a physical model.
The aircraft simulator to train pilots on new models of aircraft represents the use of physical
model as a means of experimentation. The mathematical models of real-life situations in
investment analysis, scheduling or inventory control can be experimented; these are known
as symbolic simulation models. These models can be either deterministic or probabilistic.
The deterministic models can provide answers to ‘what if’ type of questions in business
problems. The probabilistic simulation models deal with random phenomenon and the
method of simulation applied is known as Monte Carlo simulation.
16.3 MONTE CARLO SIMULATION
The Monte Carlo method is the earliest method of simulation; the method employs random
numbers and is used to solve problems that depend upon probability, where physical
experimentation is impracticable and the creation of a mathematical formula impossible.
It is method of Simulation by the sampling technique. That is, first of all, the probability
distribution of the variable under consideration is determined; then a set of random numbers
is used to generate a set of values that have the same distributional characteristics as the
actual experience it is deviced to simulate. The steps involved in carrying out Monte Carlo
Simulation are:
(i)
Select the measure of effectiveness of the problem, that is, what element is used to
measure success in improving the system modelled. This is the element one wants to
maximise or minimise. For example, this might be idle time of a service facility, or
inventory shortages per period etc.
Simulation
16.3
(ii) Identify the variables which influence the measure of effectiveness significantly. For
example, the number of service facilities in operation or the number of units in inventory
and so on.
(iii) Determine the proper cumulative probability distribution of each variable selected under
step (ii). Plot these, with the probability on the vertical axis and the values of variables
on horizontal axis.
(iv) Get a set of random numbers.
(v) Consider each random number as a decimal value of the cumulative probability
distribution. With the decimal, enter the cumulative distribution plot from the vertical
axis. Project this point horizontally, until it intersects cumulative probability distribution
curve. Then project the point of intersection down into the vertical axis.
(vi) Record the value (or values if several variables are being simulated) generated in
step (v) into the formula derived from the chosen measure of effectiveness. Solve and
record the value. This value is the measure of effectiveness for that simulated value.
(vii)Repeat steps (v) and (vi) until sample is large enough for the satisfaction of the decision
maker.
In assigning a set of random numbers, we have to decide on the entire range of random
numbers. If the cumulative probabilities are in two digits the range of random numbers to
be assigned are 00 to 99; and if in three digits, the range is from 000 to 999, and so on.
In view of the enormous computations involved, computer is usually a necessary adjunct
though below we shall deal with quite a few simple simulation problems towards its
exposition.
Example : (For finding the value π experimentally by simulation):
In the figure below is shown the arc of a circle of a unit radius in the first quadrant. Also
shown is a square OABC of side of one unit.
y
A
B
P
r=
1
y
O
x
C
Fig. 1
x
16.4
Advanced Management Accounting
The equation of the circle is given by x 2 + y2 = 1. Two* random numbers, each less than
unity (viz., x = 0.1906 and y = 0.3698) are picked up and the point P corresponding to these
is shown plotted. Obviously, if x2 + y2 < 1, P is inside or on the circle but if x2 + y2 > 1, P is
beyond the circle but within the square.
In this manner, hundreds or thousands of pairs of random numbers are picked up and it is
ascertained if the points corresponding to them fall in/on the arc or beyond the square.
Suppose that n out of the total of N points fall in/on the arc.
Now the area enclosed by the arc =
π 2 π
1 = and the area enclosed by the square = 12=1
4
4
π/4 n
4n
=
or π =
1
N
N
The experiment value of π is thus obtained. Obviously larger the sample size N, closer
shall be the true value of π.
This way, the Monte Carlo methods can be applied to solve complex and stochastic/
deterministic queuing, inventory control, production scheduling, etc. problems. The methods
have since recently been also applied in moon landing and studying galactic collisions,
atomic behavior and in military.
Illustration
Nine villages in a certain administrative area contain 720, 130, 150, 240, 960, 100, 52, 35,
532 fields respectively. Make a random selection of six fields using the random number
tables.
Solution
*
1
Village No.
2
No. of Villages
3
Cumulative No.
Of Villages
4
Random nos. fitted
against the village
As per column 3.
1
2
3
4
5
6
7
8
9
720
130
150
240
960
100
52
36
532
720
850
1000
1240
2700
2300
2352
2388
2920
0091,0684
0839,0814
2377
2471
See the Appendix on an excerpt of random nos. at the end of the chapter. How to use it is explained by example
below it. However, the nos. we are using have been taken from the other excerpt.
Simulation
16.5
The first random no. picked up from the random no. tables is 2377. Since it is immediately
< 2388 in Col. 3; it is fitted against village 8 in Col. 4. The next random no. 5997 is > 2920
in Col. 4; therefore, it is dropped. In this manner, the following random nos. are, either fitted
in Col. 4 or dropped; 8269 (D= drop), 8385 (D), 6198 (D), 0091 (Fl) = fitted against village
no. 1 col. 4)
4829 (D), 3322 (D), 0684 (Fl), 3267 (D), 8209 (D), 5166 (D), 0839 (F2),
0814 (F2), 7409 (D), 2471 (F9). We stop here because 6 fields have been selected, two
each from village nos. I and 2 and one each from village nos. 8 and 9.
Illustration
Frontier Bakery keeps stock of a popular brand of cake. Daily demand based on past
experience is as given below:Experience indicates
Daily demand
:
0
15
25
35
45
50
Probability
:
.01
.15
.20
.50
.12
.02
Consider the following sequence of random numbers:R. No. 48, 78, 09, 51, 56, 77, 15, 14, 68, 09
Using the sequence, simulate the demand for the next 10 days.
Find out the stock situation if the owner of the bakery decides to make 35 cakes every day.
Also estimate the daily average demand for the cakes on the basis of simulated data.
Solution
According to the given distribution of demand, the random number coding for various
demand levels is shown in Table below :
Random Number Coding
Demand
Probability
Cum. Prob.
Random nos. fitted
0
0.01
0.0
00
15
0.15
0.16
01-15
25
0.20
0.36
16-35
35
0.50
0.86
36-85
45
0.12
0.98
86-97
50
0.02
1.00
98-99
The simulated demand for the cakes for the next 10 days is given in the Table below. Also
given in the table is the stock situation for various days in accordance with the bakery
decision of making 35 cakes per day.
16.6
Advanced Management Accounting
Determination of Demand and Stock Levels
Day
Random Number
Demand
1
48
35
2
78
35
–
3
09
15
20
4
51
35
20
5
56
35
20
6
77
35
20
7
15
15
40
8
14
15
60
9
68
35
60
10
09
15
80
Expected demand = 270/10 =
Stock
27 units per day.
Illustration
A company manufactures around 200 mopeds. Depending upon the availability of raw
materials and other conditions, the daily production has been varying from 196 mopped to
204 mopped, whose probability distribution is as given below:Production per day
Probability
196
0.05
197
0.09
198
0.12
199
0.14
200
0.20
201
0.15
202
0.11
203
0.08
204
0.06
The finished mopeds are transported in a specially designed three storeyed lorry that can
accommodate only 200 mopped. Using the following 15 random numbers 82, 89, 78, 24,
53, 61, 18, 45, 04, 23, 50, 77, 27, 54, 10, simulate the process to find out:
(i)
What will be the average number of mopeds waiting in the factory?
(ii)
What will be the average number of empty spaces on the lorry?
Simulation
16.7
Solution
The random numbers are established as in Table below:
Production
Probability
Cumulative
Random Number
Per day
Probability
196
0.05
0.05
00-04
197
0.09
0.14
05-13
198
0.12
0.26
14-25
199
0.14
0.40
26-39
200
0.20
0.60
40-59
201
0.15
0.75
60-74
202
0.11
0.86
75-85
203
0.08
0.94
86-93
204
0.06
1.00
94-99
Based on the 15 random numbers given we simulate the production per day as above in
table 2 below.
Random No.
Estimated
Production
Per day
No. of mopeds waiting
Opening
Balance
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
82
89
78
24
53
61
18
45
04
23
50
77
27
54
10
202
203
202
198
200
201
198
200
196
198
200
202
199
200
197
–
2
5
7
5
5
6
4
4
0
0
0
2
1
1
Current
Excess
production
2
3
2
–
–
1
–
–
–
–
–
2
–
–
–
No. of empty
spaces in the lorry
Current
short
production
–
–
–
2
–
–
2
–
4
2
–
–
1
–
3
Total
waiting
2
5
7
5
5
6
4
4
0
0
–
2
1
1
–
–
–
–
–
–
–
–
–
–
2
–
–
–
–
2
Total
42
4
16.8
Advanced Management Accounting
Average number of mopeds waiting
42
=
=
2.80
=
0.266
15
Average number of empty spaces in lorry
4
=
15
(Note : Some of the authors have solved this problem without adjusting excess/short production. However, we feel that above approach in better.)
Illustration
Ramu and Raju are workers on a two-station assembly line. The distribution of activity
times at their stations is as follows:–
Time in
Time frequency
Time frequency
Sec.
For Ramu
for Raju
10
4
4
20
6
5
30
10
6
40
20
7
50
40
10
60
11
8
70
5
6
80
4
4
(a) Simulate operation of the line for eight times. Use the random numbers given below:
Operation 1
Operation 2
14
61
36
97
01
82
76
41
96
00
55
56
44
03
25
34
(b) Assuming Raju must wait until Ramu completes the first item before starting work, will
he have to wait to process any of the other eight items? Explain your answer, based upon
your simulation.
Solution
Cumulative frequency distribution for Ramu is derived below. Also fitted against it are the
eight given random numbers. In parentheses are shown the serial numbers. of random
numbers.
Simulation
10
20
30
40
50
60
70
80
4
10
20
40
80
91
96
100
01(2)
00(7)
16.9
03(8)
14(l)
44(4)
82(6)
95(3)
61(5)
Thus the eight times are: 30, 10, 70, 50, 50, 60, 10 and 10 respectively.
Likewise the eight times for Raju are derived from his cumulative distribution below:1
2
Frequency
3
Cumulative
4
2 × col. 3
10
20
30
40
4
5
6
7
4
9
15
22
8
18
30
44
50
60
70
80
10
8
6
4
32
40
46
50
64
80
92
100
13(7)
25(4)
36(1),
41(6),
55(3)
76(2)
34(8)
97(5)
(Note that cumulative frequency has been multiplied by 2 in Co. 4 in order that all the given
random numbers are utilised).
Thus, Raju’s times are 40,60,50,30,80,40,20 and 40 seconds respectively.
Ramu’s and Raju’s times are displayed below to observe for waiting time, if any.
1
Ramu
2
Cum. time
30
10
70
50
50
60
10
10
30
40
110
160
210
270
280
290
3
Raju
initial
40
60
50
30
80
40
20
40
4
Raju’s cumulative time
with 30 seconds included
70
130
180
210
290
330
350
390
Since col. 4 is consistently greater than col. 2 no subsequent waiting is involved.
16.10
Advanced Management Accounting
Illustration
Dr. STRONG is a dentist who schedules all her patients for 30 minutes appointments.
Some of the patients take more or less than 30 minutes depending on the type of dental
work to be done. The following summary shows the various categories of work, their
probabilities and the time needcd to complete the work :
Category
Time required
Probability of category
Filling
45 minutes
0.40
Crown
60 minutes
0.15
Cleaning
15 minutes
0.15
Extraction
45 minutes
0.10
Checkup
15 minutes
0.20
Simulate the dentist’s clinic for four hours and determine the average waiting time for the
patients as well as the idleness of the doctor. Assume that all the patients show up at the
clinic at exactly their scheduled arrival time starting at 8.00 a.m. Use the following random
numbers for handling the above problem:
40
82
11
34
25
66
17
79
Solution
If the numbers 00-99 are allocated in proportion to the probabilities associated with each
category of work, then various kinds of dental work can be sampled, using random number
table:
Type
Probability
Random Numbers
Filling
0.40
00-39
Crown
0.15
40-54
Cleaning
0.15
55-69
Extraction
0.10
70-79
Checkup
0.20
80-99
Using the given random numbers, a work sheet can now be completed as shown on next
page:
Future Events
Patient
1
2
3
4
5
6
7
8
Scheduled Arrival
8.00
8.30
9.00
9.30
10.00
10.30
11.00
11.30
R. No.
40
82
11
34
25
66
17
79
Category
Crown
Checkup
Filling
Filling
Filling
Cleaning
Filling
Extraction
Service Time
60 minutes
15 minutes
45 minutes
45 minutes
45 minutes
15 minutes
45 minutes
45 minutes
Simulation
16.11
Now, let us simulate the dentist’s clinic for four hours starting at 8.00 A.M.
Status
Time
Event
8.00
8.30
9.00
9.00
9.15
9.30
10.00
lst patient arrives
2nd arrives
lst departs
3rd arrives
2nd departs
4th arrives
3rd depart
5th arrives
6th arrives
4th departs
7th arrives
5th departs
8th arrives
6th departs
End
–
10.30
10.45
11.00
11.30
11.45
12.00
12.30
Number of the patient
Being served (time to go)
lst (60)
lst (30)
Patients
waiting
–
2nd
2nd(15)
3rd(45)
3rd (30)
3rd
4th
4th
4th
5th
5th
(45)
(15)
(45)
(30)
5th
5th &6th
6th
6th & 7th
6th
7th
7th
8th
(15)
(45)
(30)
(45)
7th & 8th
8th
8th
–
The dentist was not idle during the entire simulated period:
The waiting times for the patients were as follows:
Patient
Arrival
Service Starts
1
2
3
4
5
6
7
8
8.00
8.30
9.00
9.30
10.00
10.30
11.00
11.30
8.00
9.00
9.15
10.00
10.45
11.30
1l.45
12.30
Waiting
(Minutes)
0
30
15
30
45
60
45
60
Total
The average waiting time of a patient was
=
285
= 35.625 minutes.
8
285
16.12
Advanced Management Accounting
Illustration
A bakery chain delivers cakes to one of its retail stores each day. The number of cakes
delivered each day is not constant but has the following distribution:Cakes delivered per day
Probability
10
0.05
11
0.10
12
0.15
13
0.35
14
0.20
15
0.10
16
0.05
The number of customers desiring cakes each day has the distribution:
No. of customers
Probability
5
0.10
6
0.15
7
0.20
8
0.40
9
0.10
10
0.05
Finally, the probability that a customer in need of cakes wants 1, 2, or 3 cakes is described
by
Cakes to a customer
Probability
1
0.40
2
0.40
3
0.20
Estimate by Monte Carlo methods the average number of cakes left over per day and the
average number of sales per day owing to lack of cakes. Assume that left over cakes are
given away at the end of each day.
Simulation
16.13
Solution
The first cumulative probability distribution is derived below:
Cumulative probability distribution of cakes delivered per day
Cakes per day
Cumulative
Ten random
Probability
nos. fitted in
10
0.05
.0153(9)
11
0.15
.1342(6)
12
0.30
.2291(4)
13
0.65
.5878(2),.4391(5),.5210(7),.3411(8)
14
0.85
.8136(1),.7923(10).
15
0.95
16
1.00
.9655(3)
Thus, the cakes delivered in each of the 10 days are: 14, 13, 16, 12, 13, 1l, 13, 13, 10, 14.
Likewise the no. of customers per day are derived below from the cumulative probability
distribution.
No. of customers
Cumulative probability
Ten random nos. fitting in
5
0.10
.0906(9)
6
0.25
.2416(4),.1934(10)
7
.3501(l),.3396(3),.2587(5),.3072(6)
0.45
8
0.85
.5054(2)
9
0.95
.8511(7), .8698(8)
10
1.00
Thus the number of customers for the ten days are 7, 8, 7, 6, 7, 7, 9, 9, 5, 6, and the total
no. of customers = 71.
16.14
Advanced Management Accounting
Thus cakes delivered per day and the no. of customers per day are laid out below.
Days
No. of Customers
desiring cake on nth Day
No. of
cakes desired
by each customer
1
2
3
4
5
6
7
8
9
10
7
8
7
6
7
7
9
9
5
6
13
16
12
13
11
13
13
10
14
No. of cakes wanted
1
2
3
3
2
1
2
14
No. of cakes delivered
14
Lift overs
NIL
Shortage
NIL
This table has to be filled in by selecting no. of cakes demanded by each customer from
the third cumulative distribution derived below. In the first day, there are seven customers.
Thus 7 random nos. have been fitted in the cumulative probability distribution below. These,
in turn, have been put under day 1 of the above table. The student may fill in the remaining
columns himself, as an exercise.
Cumulative probability distribution of the no. of cakes wanted by the customers.
(7 random nos. fitted in below give the entries under day of 1 for the above table.)
No. of Cakes
Cumulative probability
7 random numbers fitted in for day 1
1
0.40
.23(1), .00(6)
2
0.80
.49(2), .61(5), .48(7)
3
1.00
.82(3), .84(4)
Thus the no. of cakes wanted by the 7 customers in the first day are 1, 2, 3, 3, 2, 1, 2. These
are entered under day 1. Left-overs and shortages are also computed which may be done
by student for the remaining columns.
16.4 SIMULATION AND INVENTORY CONTROL
The Monte Carlo simulation is widely used to solve inventory problems characterised by
uncertainty of demand and lead time. The distribution of demand during the lead time can
be obtained from an empirical analysis of past data or by computer simulation using random
numbers. The cumulative probability distribution of demand during the lead time is used as
Simulation
16.15
a basis to determine the annual inventory costs and stock-out costs for different levels of the
safety stock. The management can experiment the effect of various inventory policies by
using simulation and finally select an optimum inventory policy. The purpose of simulation,
as applied to inventory, is to facilitate the management in selecting an inventory policy that
will result in minimum annual inventory costs-ordering, carrying and stock-out costs. The
application of simulation to inventory control is explained below with the help of an example.
The distribution of demand during the lead time and the distribution of the lead time are set
out in Tables 1 and 2.
Table 1 : Distribution of Demand during Lead time
Quantity demanded
during lead time
0
1
2
3
Probability
0.10
0.45
0.30
0.15
Table 2 : Distribution of Lead time
Lead time (weeks)
2
3
4
Probability
0.20
0.65
0.15
Suppose, the management wants to ascertain the annual inventory costs, if they wish to
reorder when the quantity on hand is 6 units and order each time a quantity of 12 units.
Assume that all orders are delivered for the entire quantity. The cost of ordering is
Rs. 120, and the cost of holding the inventory in stock is Rs. 5 per unit per week. Further,
the management has found that when it runs out-of-stock, it costs Rs. 75 per unit.
Table 5 illustrates the manual simulation for 15 weeks. Before proceeding with simulating
the demand for each week, we calculate first the cumulative probability and assign random
numbers for each value of the two variables-demand and lead time. These are set out in
Tables 3 and 4.
Table 3
Demand
Cumulative
Random No.s
Probability
assigned
0
0.10
00 to 09
1
0.55
10 to 54
2
0.85
55 to 84
3
1.00
85 to 99
16.16
Advanced Management Accounting
Table 4
Lead Title
Cumulative
Probability
0.20
0.85
1.00
2
3
4
Random Nos.
assigned
00 to 19
20 to 84
85 to 99
We assume that the stock on hand at the start of the simulation process is 10 units. Further,
we also assume that all orders are placed at the beginning of the week and all deliveries
against orders are received at the beginning of the week.
The simulated demand for work 1 is 1 unit (corresponding to the random number 49 in
Table 3), and the closing inventory is 9 units. The inventory-carrying cost works out to
9 × 5 = Rs. 45 (Rs. 5 per unit per week). The inventory-carrying costs are calculated for the
remaining weeks in the same way. At the end of week 4, the closing inventory is 6 units
(reorder point). So an order is placed at the beginning of week 5 for a quantity of 12 units
(ordering quantity). The simulated lead time is 3 weeks (corresponding to random number
84 in Table 4). This order quantity is, therefore, received at the beginning of week 8. We
notice from Table 5 that another order is placed at the beginning of week 12, when the
stock on hand is 6 units, which is equal to the reorder point. Again, the simulated lead time
is 3 weeks. Before this quantity is received At the beginning of week 15, we find that the
quantity available at the beginning of week 14 is 1 unit and the simulated demand for that
week is 2 units, giving rise to a stock out of 1 unit. The stock-out cost is Rs. 75.
Table 5. Simulation of Demand and Lead Time for 15 weeks.
Week
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Stock
on hand
beginning
of week
Demand
Random
No.
10
9
7
7
6
3
3
2
12
10
8
6
4
1
–
49
67
06
30
95
01
10
70
80
66
69
76
86
56
84
Quantity
received
Quantity
demanded
1
2
0
1
3
0
1
2
2
2
2
2
3
2
2
–
–
–
–
–
–
–
12
–
–
–
–
–
–
12
Stock
on hand
end of
week
Inventory
carrying
costs
9
7
7
6
3
3
2
12
10
8
6
4
1
–
10
45
35
35
30
15
15
10
60
50
40
30
20
5
–
50
Stock out
Quantity
Costs
Lead Time
Random Lead
No. Time
period
84 3
79 3
1
75
Simulation
16.17
For the simulated period of 15 weeks, the total inventory costs are:Inventory carrying costs
=
Rs. 440
Ordering costs (2 orders × 20)
=
Rs. 240
Stock-out costs (1 stock-out)
=
Rs. 75
Total
=
Rs. 735
By simulating over a period of 2 to 3 years (i.e. 100 or 150 weeks), we can obtain a more
accurate picture of the annual inventory costs.
By varying the values of the variables (the ordering quantity and the reorder point), the
management can find out the effect of such a policy in terms of the annual inventory costs.
In other words, simulation permits the management to evaluate the effects of alternate
inventory policies. Also, if there is a change in the ordering, stock-out and inventory-carrying
costs, their impact on the annual inventory costs can be determined by using simulation.
16.5
MISCELLANEOUS ILLUSTRATIONS
Illustration
A company manufactures 30 items per day. The sale of these items depends upon demand
which has the following distribution:
Sales (Units)
Probability
27
0.10
28
0.15
29
0.20
30
0.35
31
0.15
32
0. 05
The production cost and sale price of each unit are Rs. 40 and Rs. 50 respectively. Any
unsold product is to be disposed off at a loss of Rs. 15 per unit. There is a penalty of Rs.
5 per unit if the demand is not met.
Using the following random numbers estimate total profit / loss for the company for the next
10 days: 10, 99, 65, 99, 95, 01, 79, 11, 16, 20
If the company decides to produce 29 items per day, what is the advantage or disadvantage
to the company?
Solution
First of all, random numbers 00-99 are allocated in proportion to the probabilities associated
with the sale of the items as given below:
16.18
Advanced Management Accounting
Sales (units)
Probability
27
28
29
30
31
32
0.10
0.15
0.20
0.35
0.15
0.05
Cumulative
probability
0.10
0.25
0.45
0.80
0.95
1.00
Random numbers
assigned
00-09
10-24
25-44
45-79
80-94
95–99
Let us now simulate the demand for next 10 days using the given number in order to
estimate the total profit/loss for the company. Since the production cost of each item is Rs.
40 and sale price is Rs. 50, therefore the profit per unit of the sold item will be Rs. 10.
There is a loss of Rs. 15 per unit associated with each unsold unit and a penalty of Rs. 5
per unit if the demand is not met. Accordingly, the profit / loss for next ten days are calculated
in column (iv ) of the table below if the company manufactures 30 items per day.
(i)
Day
1
2
3
4
5
6
7
8
9
10
(ii)
Random
number
10
99
65
99
95
01
79
11
16
20
Total Profit =
(iii)
Estimated
sale
28
32
30
32
32
27
30
28
28
28
(iv)
Profit/Loss per day when
production = 30 items per day
28 × Rs. 10 – 2 × Rs. 1.5=250
30 × Rs. 10 – 2 × Rs.5=290
30 × Rs. 1 0=300
30 × Rs. 10 – 2 × Rs.5=290
30 × Rs. 10 – 2 × Rs.5=290
27 × Rs. 10 – 3 × Rs. 1 5=225
30 × Rs.10 = 300
28 × Rs. 10 – 2 × Rs. 1 5=250
28 × Rs.10 – 2 × Rs.15=250
28 × Rs. 10 – 2 × Rs. 1 5=250
(v)
Profit/loss per day when
production = 29 items per day
28 × Rs. 10 – 1 × Rs. 15=265
29 × Rs. 10 – 3 × Rs. 5 = 275
29 × Rs.10 – 1 × Rs.5=285
29 × Rs. 10 – 3 × Rs.5=275
29 × Rs. 10 – 3 × Rs.5=275
27 × Rs. 10 – 2 × Rs. 5=240
29 × Rs. 10 – 1 × Rs. 5=285
28 × Rs. 1 0 -1 × Rs. 15=265
28 × Rs.10 - 1× Rs.15= 265
28 × Rs. 1 0 - 1 × Rs. 1 5=265
Rs.2695
Rs.2695
The total profit for next 10 days will be Rs. 2695 if the company manufactures 30 items per day.
In case, the company decides to produce 29 items per day, then the profit of the company for
next 10 days is calculated in column (v) of the above table. It is evident from this table that there
is no additional profit or loss if the production is reduced to 29 items per day since the total profit
remains unchanged i.e. Rs. 2695.
Illustration
An Investment Corporation wants to study the investment projects based on three factors : market
demand in units, price per unit minus cost per unit, and the investment required. These factors
are felt to be independent of each other. In analysing a new consumer product, the corporation
estimates the following probability distributions:
Simulation
Annual
(Price-Cost)
demand
Units
25,000
30,000
35,000
40,000
45,000
50,000
55,000
per unit
Rs.
3.00
5.00
7.00
9.00
10.00
Probability
0.05
0.10
0.20
0.30
0.20
0.10
0.05
Probability
0.10
0.20
0.40
0.20
0.10
16.19
Investment
Required
Rs.
Probability
27,50,000
30,00,000
35,00,000
0.25
0.50
0.25
Using simulation process, repeat the trial 10 times, compute the return on investment for each
trial, taking these three factors into account. Approximately, what is the most likely return? Use
the following random numbers for annual demand, (price-cost) and the investment required:
28, 57, 60, 17, 64,
20, 27, 58, 61, 30; 19, 07, 90, 02 57,
28,
29,
83,
58,
41,
18,
67,
16,
71,
43,
68,
47,
24,
19,
97
Solution
The yearly return can be determined by the formula:
Return (R) =
(Price – Cost) × Number of units demanded
Investment
The results of the simulation are shown in the table given below:
Trials
1
2
3
4
5
6
7
8
9
10
Random
Number of
Demand
28
57
60
17
64
20
27
58
61
30
Simulated
Demand
(‘000)
35
40
40
35
40
35
35
40
40
35
Random
Number
for profit
(Price-Cost)
per unit
19
07
90
02
57
28
29
83
58
41
Simulated
Profit
5.00
3.00
10.00
3.00
7.00
5.00
5.00
9.00
7.00
7.00
Random
number
for
investment
18
67
16
71
43
68
47
24
19
97
Simulated
investment
(‘000)
2750
3000
2750
3000
3000
3000
3000
2750
2750
3500
Simulated
Return (%)
Demand ×
Profit per
Investment
6.36
4.00
14.55
3.50
9.33
5.83
5.83
13.10
10.18
7.00
16.20
Advanced Management Accounting
Result: Above table shows that the highest likely return is 14.6% which is corresponding to
the annual demand of 40,000 units resulting a profit of Rs. 10 per unit and the required
investment will be Rs. 27,50,000.
Illustration
The occurrence of rain in a city on a day is dependent upon whether or not it rained on the
previous day. If it rained on the previous day, the rain distribution is given by:
Event
Probability
No rain
0.50
1 cm. rain
0.25
2 cm. rain
0.15
3 cm. rain
0.05
4 cm. rain
0.03
5 cm. rain
0.02
If it did not rain the previous day, the rain distribution is given by:
Event
Probability
No rain
0.75
1 cm. rain
0.15
2 cm. rain
0.06
3 cm. rain
0.04
Simulate the city’s weather for 10 days and determine by simulation the total days without rain
as well as the total rainfall during the period. Use the following random numbers:
67
63
39
55
29
78
70
06
78 76
for simulation. Assume that for the first day of the simulation it had not rained the day before.
Solution
The numbers 00-99 are allocated in proportion to the probabilities associated with each event.
If it rained on the previous day, the rain distribution and the random number allocation are
given below:
Event
Probability
Cumulative
Random
Probability
numbers
assigned
No rain
0.50
0.50
00-49
1 cm. rain
0.25
0.75
50-74
2 cm. rain
0.15
0.90
75-89
3 cm. rain
0.05
0.95
90-94
4 cm. rain
0.03
0.98
95-97
5 cm. rain
0.02
1.00
98-99
Table I : Rain on previous day
Simulation
16.21
Similarly, if it did not rain the previous day, the necessary distribution and the random
number allocation is given below:
Event
Probability
Cumulative
Random
Probability
numbers
assigned
No rain
0.75
0.75
00-74
1 cm. rain
0.15
0.90
75-89
2 cm. rain
0.06
0.96
90-95
3 cm. rain
0.04
1.00
96-99
Table 2: No rain on previous day
Let us now simulate tile rain fall for 10 days using the given random numbers. For tile first
day it is assumed that it had not rained the day before:
Day
Random
Event
Numbers
1
67
No rain
(from table 2)
2
63
No rain
(from table 2)
3
39
No rain
(from table 2)
4
55
No rain
(from table 2)
5
29
No rain
(from table 2)
6
78
1 cm. rain
(from table 2)
7
70
1 cm. rain
(from table 1)
8
06
No rain
(from table 1)
9
78
1 cm. rain
(from table 2)
10
76
2 cm. rain
(from table 1)
Hence, during the simulated period, it did not rain on 6 days out of 10 days. The total rain
fell during the period was 5 cm.
Illustration
The output of a production line is checked by an inspector for one or more of three different
types of defects, called defects A, B and C. If defect A occurs, the item is scrapped . If defect
B or C occurs, the item must be reworked. The time required to rework a B defect is 15
minutes and the time required to rework a C defect is 30 minutes. The probabilities of an
A, B and C defects are .1 5, .20 and .10 respectively. For ten items coming off the assembly
16.22
Advanced Management Accounting
line, determine tile number of items without any defects, the number scrapped and the
total minutes of rework time. Use the following random numbers,
RN for defect A
48
55
91
40
93
01
83
63
57
04
79
55
10
13
18
96
20
84
56
11
47
52
RN for defect B
47
36
57
09
RN for defect C
82
95
52
03
Solution
The probabilities of occurrence of A, B and C defects are 0.15, 0.20 and 0.I0 respectively. So,
tile numbers 00-99 are allocated in proportion to the probabilities associated with each of the
three defects
Defect A
Exists
Defect B
Random
Defect C
Exists ?
Random
Exists ?
Random
numbers
numbers
numbers
assigned
assigned
assigned
Yes
00-14
Yes
00-19
Yes
00-09
No
15-99
No
20-99
No
10-99
Let us now simulate the output of the assembly line for 10 items using the given random numbers
in order to determine the number of items without any defect, the number of items scrapped and
the total minutes of rework time required:
Item
RN for
RN for
RN for
Whether
Rework
No.
defect A
defect B
defect C
any defect
time (in
exists
minutes)
Remarks
1
48
47
82
None
–
–
2
55
36
95
None
–
–
3
91
57
18
None
–
–
4
40
04
96
B
15
–
5
93
79
20
None
–
–
6
01
55
84
A
–
Scrap
7
83
10
56
B
15
–
8
63
13
11
B
15
–
9
47
57
52
None
–
–
10
52
09
03
B,C
15+30=45
–
Simulation
16.23
During the simulated period, 5 out of the ten items had no defects, one item was scrapped and
90 minutes of total rework time was required by 3 items.
Illustration
The management of ABC company is considering the question of marketing a new product.
The fixed cost required in the project is Rs. 4,000. Three factors are uncertain viz. the selling
price, variable cost and the annual sales volume. The product has a life of only one year. The
management has the data on these three factors as under:
Selling
Price
Rs.
3
4
5
Probability
0.2
0.5
0.3
Variable
Cost
Rs.
1
2
3
Probability
Sales
volume
(Units)
2,000
3,000
5,000
0.3
0.6
0.1
Probability
0.3
0.3
0.4
Consider the following sequence of thirty random numbers:
81
32
60
04
46
31
67
25
24
10
40
02
39
68
08
98
59
16
66
90
12
64
79
31
86
68
82
89
25
11
Using the sequence (First 3 random numbers for the first trial, etc.) simulate the average profit for
the above project on the basis of 10 trials.
Solution
First of all, random numbers 00–99 are allocated in proportion to the probabilities associated
with each of the three variables as given under:
Selling Price
Rs.
Probabilities
Cumulative
Probabilities
Random numbers
assigned
3
4
5
Variable cost (Rs.)
0.2
0.5
0.3
0.2
0.7
1.0
00-19
20-69
70-99
1
2
3
Sales Volumes (Units)
2,000
3,000
5,000
0.3
0.6
0.1
0.3
0.9
1.0
00-29
30-89
90-99
0.3
0.3
0.4
0.3
0.6
1.0
00-29
30-59
60-99
16.24
Advanced Management Accounting
Let us now simulate the output of ten trials using the given random numbers in order to find the
average profit for the project:
S. No.
Random
Selling
Random
Variable
Random Sales Volume
No.
Price (Rs.)
No.
Cost (Rs.)
No.
1
81
5
32
2
60
5
2
04
3
46
2
31
3
3
67
4
25
1
24
2
4
10
3
40
2
02
2
5
39
4
68
2
08
2
6
59
4
66
2
90
5
7
12
3
64
2
79
5
8
31
4
86
2
68
5
9
82
5
89
2
25
2
10
11
3
98
3
16
2
(‘000 units’)
Profit = (Selling Price – Variable cost) × Sales Volume – Fixed cost.
Simulated profit in ten trials would be as follows:
S.No.
1
2
3
4
5
6
7
8
9
10
Profit
(Rs.5 – Rs.2) × 5,000 units – Rs. 4,000
(Rs.3 – Rs. 2) × 3,000 units – Rs. 4,000
(Rs. 4 – Re. 1) × 2,000 units – Rs. 4,000
(Rs.3 – Rs. 2) × 2,000 units – Rs. 4,000
(Rs.4 – Rs. 2) × 2,000 units – Rs. 4,000
(Rs.4 – Rs. 2) × 5,000 units – Rs. 4,000
(Rs.3 – Rs. 2) × 5,000 units – Rs. 4,000
(Rs.4 – Rs. 2) × 5,000 units – Rs. 4,000
(Rs. 5 – Rs. 2) × 2,000 units – Rs. 4,000
(Rs.3 – Rs. 3) × 2,000 units – Rs. 4,000
= Rs. 11,000
= Rs. – 1,000
= Rs. 2,000
= Rs. – 2,000
= 0
= Rs. 6,000
= Rs. 1,000
= Rs. 6,000
= Rs. 2,000
= Rs.-4,000
Total Rs.2 1,000
Therefore average profit per trial =
Rs. 21,000
= Rs. 2,100
10
Example on the use of random number table. Suppose we want five 3-digit random nos.
We can enter any where in the table e.g. the last column, first 3 digits of 5 consecutive nos
give us the answers : 413, l72, 207, 511, l72. Thus we can enter the table randomly but then
on proceed serially.
Simulation
16.25
16.6 RANDOM NUMBERS TABLE
8481
0744
5558
9371
3033
1053
4389
2158
6749
2878
0591
1025
4244
1331
8853
5059
0821
5262
1210
3642
9598
3894
3603
1121
1930
6309
4460
8371
9397
9304
5606
6693
0556
6973
4920
0132
4051
0267
0609
2593
8812
3540
9535
3174
8864
5245
5468
3951
7319
5892
5016
3447
7239
1463
1660
7320
3504
8854
3676
3447
6549
3438
9217
9032
3490
4192
4340
1746
1858
6629
5322
3173
3011
2978
4583
9158
3143
5095
5540
1468
2435
8333
7715
9299
1223
0098
3101
5612
9469
1666
9491
5985
5375
7677
4760
7341
6038
7928
4064
8731
0080
6173
2976
2164
6365
6532
4086
9534
4943
4804
2206
0546
1628
1388
2589
6331
3194
7108
9365
5775
3747
2853
6762
6313
4227
2830
5383
7273
9298
4013
8546
082
8994
4959
5208
8241
9854
5504
3149
5750
2602
0019
1239
8282
1129
0593
4511
6818
4024
6269
4376
3288
4836
2301
9054
7214
9434
1196
1406
6761
6185
2545
4524
5661
8744
5485
0118
6496
6562
3219
0363
9312
0848
5857
4120
3262
0327
1866
9076
7465
3209
7546
4245
7146
6661
0858
4488
7917
4086
5105
4100
7155
1624
6669
6205
5656
1440
4161
5401
5189
2579
6378
6134
3142
1155
8972
0136
4941
8614
5309
6188
1089
0163
0472
1221
5922
4773
2570
0269
8801
5995
2498
8316
8457
6893
9809
9668
4193
1299
0861
4802
2910
1540
1426
4907
7627
6931
7928
8911
4287
4962
1471
5378
5879
4949
6799
2135
4840
2834
2071
8293
6704
5120
3866
8844
6466
5695
2697
2060
0636
2649
1280
9895
7128
4667
0982
8668
4243
9923
4047
5504
8878
3485
1395
7005
4606
1697
4163
0669
6787
6690
8553
8150
7086
1102
4174
3266
8034
8547
4380
1037
185l
6803
7874
4205
3071
5777
1392
7518
4084
5950
0966
1541
8707
4085
1217
6876
7497
6433
0522
2389
6701
9586
1902
8396
9390
1891
5029
1796
6861
6804
4956
6388
7240
2264
8669
8335
2035
0088
3581
8527
8725
3889
8743
0501
1170
3147
9989
3518
7860
9778
8682
7177
9931
0222
1751
9903
1323
3978
9789
1048
9986
6514
9980
9589
0100
7937
1149
1660
2004
9483
9742
2083
0343
4779
8993
7989
8949
6626
0765
7033
8748
8925
8630
6067
1154
0860
2812
1809
7977
2210
1237
5273
6982
3625
3142
2500
4353
9349
7792
1778
9957
7911
6581
7479
4339
1048
5014
3742
3620
0650
5202
3581
2011
2924
1458
5476
1952
8875
3207
0924
3453
8459
7724
9937
3621
9190
4989
8591
2379
0957
6247
3790
3895
5627
5841
2188
4996
0631
8409
6706
4423
6977
5547
8513
2455
8011
2675
1836
1625
4223
3772
4660
9784
6470
5630
6726
7538
1005
9819
4135
1728
2079
5111
1725
9416
3396
3482
3194
4684
1389
5573
0644
0528
7206
2602
2296
5008
1990
3494
9556
6673
0898
6348
7985
7695
1908
7142
3162
4708
1930
1570
2703
3394
3192
4891
8848
9497
0548
7659
3193
4706
5669
2607
8194
9496
8602
9854
4619
4548
16.26
Advanced Management Accounting
SELF-EXAMINATION QUESTIONS
(ii) What will be the average number of empty spaces on the lorry ?
1. Define a Simulation model. Distinguish between deterministic and stochastic simulation
lll
model.
2. Discuss Monte Carlo simulation. Illustrate how would you use it in situations of
(i) Queuing and (ii) Inventory Control.
3. Explain in being advantages and disadvantages of using simulation technique.
4. A retailer deals in a perishable commodity. The daily demand and supply are variables. The
data for the past 500 days show the following demand and supply.
Supply
Availability (kg)
No. of days
10
20
30
40
50
40
50
190
150
70
Demand
Demand (kg)
No. of days
10
20
30
40
50
50
110
200
100
40
The retailer buys the commodity at Rs. 20 per kg and sells it at Rs. 30 per kg. Any commodity
remains at the end of the day, has no saleable value. Moreover, the loss (unearned profit) on
any unsatisfied demand is Rs. 8 per kg. Given the following pair of random numbers, simulate
6 days sales, demand and profit.
(31,18); (63,84); (15,79); (07,32); (43,75); (81,27)
The first random number in the pair is for supply and the second random number is for
demand viz. in the first pair (31,18) use 31 to simulate supply and 18 to simulate demand.
5. A company manufactures around 200 mopeds. Depending upon the availability of raw
materials and other conditions, the daily production has been varying from 196 mopeds to
204 mopeds whose probability distribution is as given below :
Production per day
Probability
196
197
198
199
200
201
202
203
204
0.05
0.09
0.12
0.14
0.20
0.15
0.11
0.08
0.06
Simulation
16.27
The finished mopeds are transported in a specially designed three storeyed lorry that can
accommodate only 200 mopeds. Using the following 15 random numbers 82, 89, 78, 24,
53, 61, 18, 45, 04, 23, 50, 77, 27, 54, 10 simulate the process to find out :
(i) What will be the average number of mopeds, waiting in the factory ?
(ii) What will be the average number of empty spaces on the lorry ?